## Some calculations of sigma_2

On four-manifold ${(M^4,g_0)}$, we define Shouten tensor

$\displaystyle A = Ric-\frac 16 Rg$

and Einstein tensor and gravitational tensor

$\displaystyle E=Ric - \frac 14 Rg\quad S=-Ric+\frac{1}{2}Rg$

Suppose ${\sigma_2}$ is the elemantary symmetric function

$\displaystyle \sigma_2(\lambda)=\sum_{i

Thinking of ${A}$ as a tensor of type ${(1,1)}$. ${\sigma_2(A)}$ is defined as ${\sigma_2}$ applied to eigenvalues of ${A}$. Then

$\displaystyle \sigma_2(A)= \frac{1}{2}[(tr_g A)^2-\langle A, A\rangle_g] \ \ \ \ \ (1)$

Notice ${A=E+\frac{1}{12}Rg}$. Easy calculation reveals that

$\displaystyle \sigma_2(A)=-\frac{1}{2}|E|^2+\frac{1}{24}R^2 \ \ \ \ \ (2)$

Under conformal change of metric ${g=e^{2w}g_0}$, we have

$\displaystyle R= e^{-2w}(R_0-6\Delta_0 w-6|\nabla_0 w|^2) \ \ \ \ \ (3)$

$\displaystyle A=A_0-2\nabla^2_0 w+2dw\otimes dw-|\nabla_0w|^2g_0 \ \ \ \ \ (4)$

$\displaystyle S=S_0+2\nabla_0^2w-2\Delta_0wg_0-2dw\otimes dw-|\nabla_0 w|^2g_0 \ \ \ \ \ (5)$

We want to solve the equation ${\sigma_2(A)=f>0}$, which is equivalent to solve

$\displaystyle \sigma_2(A_0-2\nabla^2_0w+2dw\otimes dw-|\nabla_0w|^2g_0)=f$

This is an fully nonlinear equation of Monge-Ampere type. Under local coordinates, the above equation can be treated as

$\displaystyle F(\partial_i\partial_j w,\partial_kw,w,x)=f$

where ${F(p_{ij},v_k,s,x):\mathbb{R}^{n\times n}\times\mathbb{R}^n\times\mathbb{R}\times\mathbb{R}^n\rightarrow \mathbb{R}}$. This equation is elliptic if the matrix ${\left(\frac{\partial F}{\partial p_{ij}}\right)}$ is positive definite. In order to find that matrix, we need the linearized operator

$\displaystyle L[\phi]=\frac{\partial F}{\partial p_{ij}}(\nabla_0^2\phi)_{ij}=\frac{d}{dt}|_{t=0}F(\partial_i\partial_j w+t\partial_i\partial_j\phi,\partial_kw,w,x) \ \ \ \ \ (6)$

Using the elementary identity

$\displaystyle \frac{d}{dt}\rvert_{t=0}\sigma_2(H+tG)=tr_gH\cdot tr_gG-\langle H, G\rangle_g. \ \ \ \ \ (7)$

for any fixed matrix ${H}$ and ${G}$. Now plug in ${H=A}$ is Schouten tensor and ${B=-2\nabla_0^2\phi}$. One can calculate them as

$\displaystyle tr_g H\cdot tr_g G=\langle \frac{1}{3}Rg, G\rangle_g \ \ \ \ \ (8)$

Then we get

$\displaystyle L[\phi]=\langle S,G\rangle_g=-2\langle S,\nabla^2_0\phi\rangle_g$